Knowing that $\phi$ is a characteristic function, show that $e^{\phi-1}$ is also a characteristic function

Question

Knowing that $\phi$ is a characteristic function of a random variable $X$, how to show that $e^{\phi-1}$ is also a characteristic function of some random variable.

I tried this way:
$e^{\phi-1}=e^{-1}\sum_{i=0}^{\infty}\frac{\phi^k}{k!}=e^{-1}E(\phi^Y)$ where $Y$ has $Pois(1)$ distribution.

Suppose that $X_i$ iid with the same distribution as $X$,
$e^{-1}E(\phi^Y)=e^{-1}E(E(e^{\sum_{i=1}^YitX_i}))=e^{-1}E(E(E(e^{\sum_{i=1}^YitX_i}|Y)))$.

And I got stuck here, because I don't see it to prove anything.

Answer 1

$E{e^{{it\sum_1 ^{Y} X_j}}=\sum_n e^{-1} \frac 1 {n!} {\phi (t)}^{n}}=e^{\phi (t) -1}$. In the first step I assumed that Y is independent of the $X_j$'s so that I can condition on $\{Y=n\}$ and then sum over n. The last step is just summing the exponential series.

Answer 2

In fact, you have just solved the problem.

Just suppose $\xi=X_1+X_2+\cdots+X_Y$ where $Y$ follows a Poisson distribution with event rate $1$ and $\{X_i\}_{i \ge 1}$ is a family of identically independent random variables whose characteristic functions are both $\phi(t)$. Then we have $\mathbb{E} e^{it\xi} = e^{\phi-1}$.